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Exercise 14: Algebra 1.3
p.35: 1, 2, 5, 9, 13, 15, 21, 22, 25, 31, 32, 40, 45,
D3, D4
↳
:
D=
axtb
Now
the
slope
of
the
line
(
that
is
a
)
is
-1
.
(
for
example
,
we
take
two
points
#
I
4
:,,,↳h
A
(0.1121×1,41)
191
13
(
1
,
0
)
=
(
Xzcyr
)
•
•
☐
>
×
then
slope
=
=
¥0
=
-1
.
0=1%0 7
So
y=
-
✗
c-
b.
Now
we
plug
into
the
point
#
(
D.
1)
L
,
:
(
vertical
line
so
all
points
on
Li
share
the
same
✗
-
coordinat e
)
so
I
=
-
Otb
b=
,
y
✗
=
I
☐
Lz
Y
=
I
↳
:
it
passing
through
10 ,0
)
&
(
1.
1)
D=
✗
Lx
y
=
-
✗
+
I
wanminliu@gmail.com
Ftl
(b)
✗
z
↳
Ideas
;
we
can
use
a
veto
to
describe
it
.
All
point
on
↳
is
a
↳
scalar
of
the
rear
ops
OÑ=
II.
,
I
,
1)
Then
the
the
can
be
given
by
¥
IX.
7.
t
)
=
1-
(
1.1
,
1)
L
,
:
vertical
line
passing
through
Pllc
1.
1)
4
↳
AnotherWay__
✗
=/
and
9=1
.
✗
=
t
Lz
7=1
,
and
2-
=p
,
{
Y
=
t
2-
it
↳
✗
=L
and
2-
=/
.
the
origin
.
×"%!%i,
.
Ly
:
Passy
thwy
P
and
0=10,99
☐
wanminliu@gmail.com
#
5
Find
valor
and
parametric
efuatlm
Reward
.
The
choice
of
parameter
of
the
line
determined
by
the
given
points
is
NOT
unique
,
{
✗
=×
Y
=
§
✗
or
directly
Y=§×
with
parameter
×
itself
.
-
(
b)(h1,l)wdK°
¥¥*
.
5=1
.
This
is
EI
Lk
.
/
¥
?
}
with
parameter
t
.
.
Opt
=
13
,
5)
+
=t
☐
Then
the
line
Passing
through
of
(C)
Q=(
1
,
-1
,
1)
and
R=l2
,
1
.
1)
is
given
by
ThevertmQR-coorda.at#
IX.
4)
=
t
(
3
.
5)
ending
points
-
coordinate
of
{
✗
=
not
with
parameter
t
.
☐
starting
point
y=
st
=
(
2
,
I
-
1)
-
Vil
-
1)
=
(
1,2
,
☐
,
,
wanminliu@gmail.com
Then
any
point
M-tx.y.tt
)
on
the
line
passing
through
QÑ
can
be
described
as
QNT
is
Colin
onto
QÑ
QM→
=
t
apt
QT
=
(
X.Y.tl
-
(
I
,
-1
,
1)
IX.
Y
,
t
)
-
11
,
-1.1
)
=
tar
=
t.zt.io
)
µ
-1
,
Y
-11
,
2-
4)
=
Lt
,
2T
,
0
)
so
4×-1=-1
✗
=
t
-11
4-11
=
It
=)
{
Y
=
zt
-
l
z
-
I
=
°
d-
=\
☐
wanminliu@gmail.com
#
9
.
Find
a
point
-
normal
quoth
5¥
we
know
of
the
plane
that
panty
thrush
pin
is
pertandiudw
to
Ñ
the
point
P
and
has
normal
Ñ
.
Then
put
•
Ñ
=
0
.
Here
Ñ=
13,2
,
1)
☒
-11
,
Ytl
,
2-+1
)
•
f
3
.
2.
1)
=
@
•="
31×+1
)
-124+0
+
1.
(2-+1)=0
.
→
We
coma
stop
here
.
P
You
can
also
simplify
it
.
3×+24
+
Z
t
6
=
0
.
We
hone
veto
Pmt
=
1×-421-(-1.1-1)
☐
=
Ctl
,
4+1,2--11
)
wanminliu@gmail.com
#
13µg
Find
aviator
equator
of
the
line
whose
parametric
equation
are
{
✗
=
2+4
+
y
=
-
Itt
z
fx
-2
,
y
-11
.
Z
)
=
(
4T
,
t
.
t
)
we
can
also
write
it
as
IX.
Y
.
2-
)
=
I
2
,
-1
,
o
)
1-
t
I
4.1.11
=
t
(
4
,
1
.
1)
.
✗
=
2
+
4
t
for
y
=
-1
+
t
)
2-
=
t
f
✗
-2
,
y
-11
.
2-
)
=
IX.
Y
.
-2
)
-
f
2
,
-1
,
0
)
☐
I
passing
thigh
Q
,
but
in
the
direction
of
14,1
,
I
,
wanminliu@gmail.com
(b)
.
Find
a
vertueguatm
of
the
plane
whose
parametric
quarters
are
✗
=
It
zt
,
t.tt
{
Y
=
-2
-
-4
+5
ti
z
=
4
ti
-
ti
-
IX.
Y
.
2-
)
-
l
1
,
-2
,
°
)
=
t
,
I
2
,
-1
,
4)
+
ta
l
l
,
5
.
-1
)
.
☐
wanminliu@gmail.com
(C)
Find
parameter
equations
of
the
plane
3×+47-22=42
✗
=
✗
{
✗
=
y
2
=
3-2×+2
Y
-
2
So
IX.
7.
t
)
=
✗
(
1,0
,
?
)
-1710
,
1
,
2)
f-
10,0
,
-
2)
with
parameters
✗
and
y
,
☐
wanminliu@gmail.com
D3.ms#
(9)
The
set
of
al
vertus
in
or
'
thetas
orthogonal
to
a
honor
vertus
→
we
call
it
is
what
kind
of
geometric
object
?
(
a.
b)
and
(
a.
b)
£10.0
)
A
vertu
in
Ñ
can
Be
write
in
the
form
of
ops
.
(
we
put
its
initial
point
at
origin
?
And
we
had
the
coordinate
of
Ñ=(
✗
is
so
op
is
orthogmi
to
d.
b)
0J
•
I
a.
b)
=
0
.
If
,
IX.
Y
)
•
f
a.
b)
=
o
UX
.
-14=0
.
if
we
put
staring
point
at
origin
.
thntheending
points
lies
in
the
line
axtby-o.of-ohevm.tv
wanminliu@gmail.com
E
as
✗
a-
+
y
5
for
some
coefficients
(C)
.
{
The
set
of
all
vatos
i.
pi
✗
and
Y
.
that
al
orthogonal
to
{
two
non
collina
vectors
.
/
since
§
is
orthgnd
to
of
what
kind
of
geometric
objet
?
É
-
É
=
0
①
-
c-
.
I
=
o
②
let
t
and
8
be
two
non
-
collinear
waters
so
I
-1-8
.
51=5
.
①
c-
.
I
✗
a-
)
=
o
f
If
a-
=3
or
8=8
,
then
E.
I
are
collinear
②
c-
•
f
y
5)
=
☐
Add
them
together
the
Rad
8
are
non
collinear
E.
(
xñ#¥=o
so
I
and
I
will
span
the
whole
IR
?
e-
.
8=0
let
8
be
any
vector
in
the
set
venter
that
are
orthogonal
in
the
pr oblem
.
we
can
away
,
win
qÉ=]
8
veto
is
the
only
to
two
non
-
collinear
This
is
the
property
of
Inner
product
vector
,
wanminliu@gmail.com
(b)
.
The
set
of
all
vato s
in
or
}
that
are
orthogonal
to
a
nonzer o
vertu
>
is
whet
kind
of
geometric
objat.TN
¥
.
Devote
the
vector
by
08
(
i.e
we
put
the
initial
of
vertov
at
the
origin
,
with
ending
of
veto
at
point
P
?
Then
P
lies
on
the
plane
with
the
given
vertu
at
normal
vertu
☒
"
nouveau
no
TIKA
wanminliu@gmail.com
21-1
Find
parametric
equation
,
so
3×+27
-
2-
=
4
of
the
plane
that
is
parallel
to
f-
.
the
plane
z×+zy
-
z
=
,
we
can
use
y
and
2-
as
parameters
then
and
passing
through
the
P
11.1.11
.
-
✗
=
-
¥
-1¥
+
§
Method
I
Y=y
←
my
plane
should
be
of
2-
it
the
form
☒
-7
.
2-
)
=
YI
-
¥
,
I
,
0
)
+
2-
(
§
,
°
,
1)
3×+4
-
2-
=
K
.
{
+
1¥
.
0.0N
for
some
number
K
.
Sine
it
pass
through
P
,
we
can
call
it
t
,
call
it
-4
.
3.
I
+
2.
I
-
I
=
K
so
1<=4
wanminliu@gmail.com
21-1
Find
parametric
quarters
4-1,7-1.2--1
)
•
13,2
,
-
D=
?
of
the
plane
that
is
parallel
to
3×+4
-
2-
=
4
the
plane
3×+4-7=1
T
and
P
"
"
#
Mth
the
P
"
'
"
"
④
om
wt
"
"
hehe
)
-
asmeth
method
2
Tet
cx.Y.tl
be
in
the
plane
we
put
X
and
Y
as
she
the
plane
is
parallel
to
3×+4
-
t
'
'
parameter
Then
then
the
normal
vector
of
plane
is
13,2
,
-4
.
✗
=
✗
✗
=y
So
µ
-1
,
y
-1
,
2-
-1
)
is
orthgmd
2-
=
3×+4-4
to
(
3
,
2
,
-1
)
.
(
✗
'
%
2)
=
✗
II.
0,3
)
-1710
.
1.
2)
-110,0
,
-
4)
.
I
,
I
wanminliu@gmail.com
1=22-11
Find
parametric
equation
why
?
4-
the
plane
through
theokjmO_
The
original
eglinton
is
given
by
that
is
parallel
1-
the
plane
IX.
Y
.
t
)
-
I
0
,
-4
.
D=
-1,11
,
3.
4)
✗
=
ti
t
ti
+
tz
/
1
.
-1.0
?
{
Y
=
4
+
3
ti
-
ti
so
the
plane
is
in
the
z=4
direction
of
-4
11.3.41
-1%11.19
Sof
.
and
passing
through
to
.
-
4.
%
.
go
all
other
plane
that
is
parallel
to
sua
Y
=
By
+
3
ti
-
ti
plane
and
pantry
thug
a
point
(
a.
b.
c)
•
www.i-H-Y.z
)
-
la
.
b.
c)
=
t.lt
.
}
,
4)
1-
tally
;)
Now
(
a
,
b.
c)
=
10,90
)
.
wanminliu@gmail.com
#
25
Find
parametric
quarters
of
the
the
that
is
perpendicular
to
the
plane
Xtytt
=o
,
and
passes
through
the
point
plz.o.is
.
=.Éz
-
okay
.
letmtxiy
,
t
)
be
a
point
go
Ñm=
+
11.1.1
)
.
*
"
t
ontte
line
.
1×-2=-1
✗
=
zit
Y
-
o
=
t
I
.
So
pÑ=
1×-2
,
Y
-0
,
2-
-
1)
2--1
=
t
z
=
Itt
is
a
normal
vertu
of
the
plane
sine
it
is
perpendicular
to
-6
plane
.
The
venter
(
1.
1,1
)
is
also
a
normal
vertu
of
the
plane
.
wanminliu@gmail.com
31-1
Determine
whether
the
line
and
A
normal
veto
of
plane
is
plane
are
parable
,
11
.
2
,
3)
.
=
-
s
-
✗
+
>
=
☒
'
+
|
"
"
"
"
"
=
?
We
check
the
"
"
"
"
"
"
2-
=
3
-12T
(-4-1.2)
•
11.2
.
3)
time
plane
=
-4-2+6
=
0
←
How
to
fend
the
diverter
9-
the
6-
so
tie
line
is
parallel
to
☒
it
.
2)
-
1-
5,1
,
3)
=
t
(-4-1,2)
.
the
plane
.
so
a
direction
of
line
Is
given
by
(-4-1,2)
.
wanminliu@gmail.com
1=145-1
Let
P
=
12.3
,
-
2)
41
.
let
m
be
the
midpoint
with
coordinate
IX
-
Y
,
2-1
.
and
Q=
17
.
-4
.
11
a)
Find
the
midpoint
of
the
line
b.
(
pnt
=
MOT
)
Segment
connatind
P
and
Q
|pñ=IP
(b)
Find
the
point
on
the
the
segment
H
connecting
P
and
Q
that
:S
Then
1×-2
,
Y
-3.2-+4 =1/7-2
,
-4-3,1-1-4
)
¥
of
the
way
from
P
to
Q
.
So
☒
-2
,
y
-3
,
2-+4
=
(
E-
,
¥
,
£
)
solx.y.tl-l-zi-zs-zl.se
÷
:
(b)
Now
let
us
write
the
coordinate
N
by
IX.
Y
.
7
?
pN→=¥
pot
wanminliu@gmail.com
So
1×-2,7-3.2-+4=3-4/7-2
,
-4-3,1-1-4
)
✗
=
21--43. 5
=
¥
4--3+2,1-71
=
-5,9
7=-2
t
f.
(3)
=
¥
